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Solve $ \cos 1\cos 2\cos 3.........\cos 90^\circ . $

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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Hint: This type of equation involves the value of trigonometric functions at specific angles. This can be solved by either finding the value of the first term or the last term i.e by finding the value of either $ \cos 1^\circ $ or $ \cos 90^\circ $ .

Complete step-by-step answer:
To solve the question,
 $ \Rightarrow \cos 1\cos 2\cos 3.........\cos 90^\circ . $
But, we know the value of last term,
 $ \Rightarrow \cos 90^\circ = 0 $
So the equation becomes,
 $ \Rightarrow \cos 1\cos 2\cos 3......... \times \left( 0 \right) $
So the whole equation multiplied by $ 0 $ becomes,
 $ \Rightarrow 0. $
Therefore the solution of $ \cos 1\cos 2\cos 3.........\cos 90^\circ $ is $ 0. $
So, the correct answer is “0”.

Note: $ \Rightarrow $ Trigonometry is one of the important branches in the history of mathematics. Here, we will study the relationship between the sides and angles of a right-angled triangle. The basics of trigonometry define three primary functions which are sine, cosine and tangent.
 $ \Rightarrow $ So, this was one of the cosine problems, and we can see the problems like in sine and tangent angles also.
 $ \Rightarrow $ Its applications are in various fields like oceanography, seismology, meteorology, physical sciences, astronomy, acoustics, navigation, electronics, etc.
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